Problems from Calculation of volumes by direct integration

Development:

For the symmetry of the problem, we will consider cylindrical coordinates.

The cylinder remains determined by $r^2=1$ and the planes $z=0$ and $z=2-\cos \phi$. Therefore, the integration limits will be:

$\begin{array}{l}r\in (0,1)\\ \phi \in (0,2\pi )\\ z\in (0,2-r\cdot \cos \phi )\end{array}$

Then, if $V$ is the region under consideration:

$$\begin{array}{rl}\text{Vol}(V)=& {\iiint }_{V}1dxdydz={\int }_0^{2\pi }{\int }_0^1{\int }_0^{2-r\cos \phi }rdzdrd\theta \\ =& {\int }_0^{2\pi }{\int }_0^1[r\cdot z{]}_0^{2-r\cos \phi }drd\theta ={\int }_0^{2\pi }{\int }_0^1(2-r\cos \phi )rdrd\theta \\ =& {\int }_0^{2\pi }{\int }_0^1(2r-r^2\cos \phi )drd\phi ={\int }_0^{2\pi }[r^2-{\displaystyle \frac{r^3}{3}}\cos \phi {]}_0^{1}drd\phi \\ =& {\int }_0^{2\pi }1-{\displaystyle \frac{1}{3}}\cos \phi d\phi =[\phi -{\displaystyle \frac{1}{3}}\sin \phi {]}_0^{2\pi }=2\pi \end{array}$$

Solution:

$\text{Vol}(V)=2\pi$

Hide solution and development